Charge induced on parallel plates if a charge between plates

In summary: The Green method finds the charge distribution. I searched for "two infinite grounded conducting planes". It's problem 1.13 from Jackson.
  • #1
Rakesh Kumar Jaiswal
3
1
Two infinite conducting plates 1 and 2 are separated by a distance l. A point charge q is located between the plates at a distance x from plate 1. Find the charges induced on each plate.

Many of the places this problem has been solved by considering an infinitely large plate in place of the point charge. This solution is also given in the Irodov book itself. I get the answer by this assumption but I don't understand this assumption fully. I have tried a lot to find any alternate method to solve this problem but could not get one. Now I am posting here with a lot of hope that somebody would certainly help me out.


Can anybody solve this question by any alternate method?
 
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  • #2
Rakesh Kumar Jaiswal said:
Two infinite conducting plates 1 and 2 are separated by a distance l. A point charge q is located between the plates at a distance x from plate 1. Find the charges induced on each plate.

Many of the places this problem has been solved by considering an infinitely large plate in place of the point charge. This solution is also given in the Irodov book itself. I get the answer by this assumption but I don't understand this assumption fully. I have tried a lot to find any alternate method to solve this problem but could not get one. Now I am posting here with a lot of hope that somebody would certainly help me out.


Can anybody solve this question by any alternate method?

One solution would be to use the method of images. But, in this case, you will need an infinite sequence of image charges in both directions.

Using Green's reciprocation theorem is another approach.
 
  • #3
PeroK said:
One solution would be to use the method of images. But, in this case, you will need an infinite sequence of image charges in both directions.

Using Green's reciprocation theorem is another approach.

Thanks ReroK

I have tried to solve this by image method also but could not get the answer. At the limiting case (boundary conditions) I get equal charges induced on both plates.

Would you please solve this for me?
 
  • #4
Rakesh Kumar Jaiswal said:
Thanks ReroK

I have tried to solve this by image method also but could not get the answer. At the limiting case (boundary conditions) I get equal charges induced on both plates.

Would you please solve this for me?

I'm afraid, we can't just do your homework.

For what it's worth, I suspect the image method is quite hard. Quite complicated algebraically, at least.
 
  • #5
PeroK said:
I'm afraid, we can't just do your homework.

For what it's worth, I suspect the image method is quite hard. Quite complicated algebraically, at least.

Many people says this problem can be solved by image method but I could not find one who could actually solve this.

I am a physics faculty and I do my homework. It is long waited question to be solved for me. Please help me if you can.
 
  • #6
Rakesh Kumar Jaiswal said:
Many people says this problem can be solved by image method but I could not find one who could actually solve this.

I am a physics faculty and I do my homework. It is long waited question to be solved for me. Please help me if you can.

Well, I found at least one solution just by searching:

"method of images two parallel planes"

Although it seems that the method of images leads to an infinite series that cannot be simplified.

I also found a solution based on the Green reciprocation theorem, which looks like the way to go.
 
  • #7
PeroK said:
Well, I found at least one solution just by searching:

"method of images two parallel planes"

Although it seems that the method of images leads to an infinite series that cannot be simplified.

I also found a solution based on the Green reciprocation theorem, which looks like the way to go.
The question asks for
Rakesh Kumar Jaiswal said:
Find the charges induced on each plate.
I take that as meaning the induced charge distribution. Do those methods answer that or only find the field/potential?
 
  • #8
haruspex said:
The question asks for

I take that as meaning the induced charge distribution. Do those methods answer that or only find the field/potential?

The Green method finds the charge distribution. I searched for "two infinite grounded conducting planes". It's problem 1.13 from Jackson.

The method of images solution I found was looking for the force on the particle.
 

1. What is the formula for calculating the charge induced on parallel plates?

The formula for calculating the charge induced on parallel plates is Q = ε0 * A * ΔV / d, where Q is the induced charge, ε0 is the permittivity of free space, A is the area of the plates, ΔV is the potential difference between the plates, and d is the distance between the plates.

2. How does the charge on the plates change if the distance between them is increased?

If the distance between the plates is increased, the charge induced on the plates will decrease. This is because the electric field between the plates will weaken, resulting in a smaller charge being induced.

3. What happens to the charge on the plates if the potential difference between them is increased?

If the potential difference between the plates is increased, the charge induced on the plates will also increase. This is because a higher potential difference means a stronger electric field, resulting in a larger charge being induced on the plates.

4. Can the charge induced on parallel plates be negative?

Yes, the charge induced on parallel plates can be negative. This can happen if the plates have opposite charges and the induced charge on one plate is smaller than the opposite charge on the other plate. Alternatively, it can also occur if the plates have the same charge and the induced charge is smaller than the original charge on one of the plates.

5. How does the area of the plates affect the induced charge?

The area of the plates does not directly affect the induced charge. However, a larger plate area can result in a stronger electric field between the plates, which can in turn lead to a larger charge being induced on the plates.

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